Staff: Mentor

a 0.25 kg piece of ice at -30°C is warmed by an electric heater and the following graph of temperature is produced.
-use the information on the graph to determine the power output of the heater

the graph shows that the ice went from -30degC to -10 degC in 150s2. Relevant equations
Q=mcΔt
p=w/Δt

3. The attempt at a solution
Q=mcΔt
=(0.25)(3.3*105)((-30)-(-20))
=82500*(-20)
1.65*106

should this be negative?

The value you've used for the specific heat of ice looks odd to me. Where did it come from and what are the units?
Note that the temperature change was positive, rising from -30C up to -20C. You've got the order of the temperatures reversed. You should have ΔT = Tfinal - Tinitial

p=w/Δt
=1.65*106/150
=11000

You might want to use different variable names for the change in time and change in temperature. It is common to use upper case T for temperatures and lower case t for times.